Mark the correct alternative in the following:

The distance of the point (–1, –5, –10) from the point of intersection of the line and the plane is


Let, the point of intersection of the line


and the plane


be (x0, y0, z0).


As (x0, y0, z0) is the point of intersection of the line and the plane, so the position vector of this point i.e.


must satisfy both of the equation of


line and the equation of plane.


Substituting, in place of in both the equations, we


get,



And, ………………. (2)


i.e.
x0 =2 + 3λ


y0 = - 1 + 4λ


z0 =2 + 12λ


Substituting, these values in equation (2) we get,


((2 + 3λ)×1) - (1×( - 1 + 4λ)) + (1×(2 + 12λ))=5


2 + 3λ + 1 - 4λ + 2 + 12λ=5


11λ=0


λ=0


x0 =2 + 3λ


=2


y_0= - 1 + 4λ


= - 1


z_0=2 + 12λ


=2


Hence, the point of intersection is, (2, - 2, 2).


Now, the distance between the point ( - 1, - 5, - 10) and (2, - 1, 2) is,






=13


Hence, the required distance between the point ( - 1, - 5, - 10) the point where the line the plane , is 13 units.

1